（HDU

（HDU - 1358）Period

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9384 Accepted Submission(s): 4496

Problem Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as AK , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

Sample Input

3
aaa
12
aabaabaabaab
0

Sample Output

Test case #1

2 2
3 3

Test case #2

2 2
6 2
9 3
12 4

题目大意：给出一个有n个字符的字符串，请问在长度为[2,n]的字符子串中，是否存在循环节，如果有，输出子串长度和循环的次数。

思路：kmp求循环节模板。

#include<cstdio>using namespace std;const int maxn=1000005;char s[maxn];int Next[maxn];int n;void getNext(){    Next[0]=Next[1]=0;    for(int i=1;i<n;i++)    {        int j=Next[i];        while(j&&s[i]!=s[j]) j=Next[j];        if(s[i]==s[j]) Next[i+1]=j+1;        else Next[i+1]=0;    }}int main(){    int kase=0;    while(~scanf("%d",&n)&&n)    {        scanf("%s",s);        getNext();        printf("Test case #%d\n",++kase);        for(int i=1;i<=n;i++)            if(Next[i]&&(i%(i-Next[i]))==0) printf("%d %d\n",i,i/(i-Next[i]));        printf("\n");    }    return 0;}